Grade 8 Unit 1 Constructed Response Rational and Irrational Numbers Scoring Rubric Task 1. Decimal Representations/ Expansions: Converting Fractions to Decimals and Terminating/Nonterminating Decimals into Fractions
Common Core State Standard 8.NS.1 [a/s]: Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
8.NS.1 [a/s]: Know that numbers that are not 2. Estimating Irrational Numbers
3. Approximating Other Irrational Numbers and Their Expressions
rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
8.NS.2 [a/s]: Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g.,Ο 2).
Standards for Mathematical Practice SMP.1, SMP.2, SMP.3, SMP.6, SMP.7, SMP.8
SMP.1, SMP.2, SMP.3, SMP.7
SMP.1, SMP.2, SMP.6, SMP.7
Note to Teacher: The following scoring rubric should be used as a guide to determine points given to students for each question answered. Students are required to show the process through which they arrived at their answers for every question involving problem solving. For questions involving a written answer, full points should be given to answers that are written in complete sentences which address each component of the questions being asked.
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Grade 8 Unit 1 Constructed Response Rational and Irrational Numbers Scoring Rubric Question 1. a. Student gives correct answer and shows work: y = β3 b. Student gives correct answer and explanation: After analyzing my solution, I am able to conclude that the number negative three is an integer and a rational number. It is an integer because it is a whole number and a negative natural number. It is also a rational number because it is an π integer that can be written in the form π. c. Student gives correct examples: Answers may vary. Ex.: β12, ββ58 , 0.3333 β¦ 2. a. Student gives correct answer and show your work: Decimal Expansion: 3 0.375; Fraction: 8 b. Student gives correct answer and explanation: After writing the fractional and decimal representations of the statement given, I observed that the decimal form is a terminating, non-repeating decimal and rational number. It is a terminating, non-repeating decimal because after going through the process of changing the fraction into the decimal my answer was only 0.375 with no additional or repeating numbers after the value in the hundredths place. It is a rational number because it is a terminating π integer that can be written in the form π. 3. a. Student completes table and gives correct answer: ο· Mathematical Process: ο· Explanation of Steps: In order to find the solution I first set up the equation x = 2.3939. My next step was to get the portion of the decimal to the left-hand side of the decimal by multiplying the whole equation by 100. This gave me the new equation 100x = 239.3939. I then subtracted my original equation x = 2.3939 from the new equation 100x = 239.3939 in order to get the decimal on the right hand side of the equation to be a whole number. Once I subtracted the two equations my next step was to solve for x and in 237 doing so I got the arrived at my solution 99 . 237
ο· Answer: 99 b. Student gives explanation: In order to show that my answer is correct I could convert the fraction from my answer back into a decimal. If I get the decimal given to me in the problem as my answer then I know the fraction I in my answer is correct.
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Grade 8 Unit 1 Constructed Response Rational and Irrational Numbers 4. a. Student gives explanation: After reading the problem I was able to conclude that the problem is asking me to find the square of the largest perfect square that is smaller than β125 and the square of the smallest perfect square that is larger than β125. With this information I know that my starting point for solving this problem is to first find these two perfect squares and write an inequality comparing them to numbers I am working with β125. b. Student gives correct answer: The square root of 125 is between the integers 10 and 11. c. Student gives explanation: If the square root were changed to 5β125, the process to solve the problem would change because I would need to multiply the number 5 on the outside of the radical by each of the squares of the perfect squares that I found from the inequality I wrote. This would change the answer to 5β125 is between the integers 50 and 55. 5. a. Student gives correct answer and shows work: 9.42 b. Student gives explanation: A concept that I have learned before that was useful when finding my solution was the process for finding the two perfect squares that the square root of an irrational number is between. This concept was important when finding my solution because I needed these two perfect squares to find the decimal expansion that represents how far away the irrational number is from the largest perfect square, and it was also added to the smallest perfect square for my final answer. 6. a. Student gives correct answer and explanation: After analyzing the data given in the statement I was able to conclude that the statement is 6 incorrect. The mixed number 2 9 is not larger than β10. I know this
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because I converted the mixed number 2 9 to itβs decimal expansion and I got 2.66. When I estimated the value of β10 I got 3.1. After comparing these two decimal expansions I was able to conclude that the statement 6 was incorrect and that β10 is actually larger than the mixed number 2 9 b. Student creates number line: 6
2 9 = 2.66
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Grade 8 Unit 1 Constructed Response Rational and Irrational Numbers 7. a. Student gives correct answer and shows work: 2π β 2π 2 + π 3 b. Student gives explanation: After analyzing the expression I wrote above, I know that to calculate the value of the expression I will need to substitute the irrational numbers with their rational expressions. In this case the irrational number I am working with is the value of π. To facilitate my process for finding the value of the whole expression I am going to substitute π with an approximate value of 3.14. After substituting this value in the equation I can simplify it and solve. c. Student gives correct answer: 17.52 8. a. Student gives correct answer: 6β51 + 6β15 β π b. Student gives explanation: A mathematical concept that I learned before that was important when solving this problem was combining like terms to simplify an expression. The radicals in this expression were similar to working with variables because only like radicals could be combined. For example in the problem I was able to subtract 8β51 and 2β51 because they both have a common radical of β51, and I was also able to add 3β15 and 3β15 because of the common radical β15. c. Student gives correct answer and shows work: d. Student gives explanation: It is a good strategy to first simplify this type of expression because when you combine like terms you are making the expression smaller and easier to work with. If we did not simplify it, we would be approximating the value of every term in the expression instead of finding the approximation of fewer combined like terms. 9. a. Student gives correct answer and shows work: 326.14 b. Student gives explanation: After calculating the value of the expression I was able to determine certain differences and similarities between the process process to solve this problem and others was that I was working with square roots to the 2nd or 3rd power. This created and extra step that had to be done. This problem was similar to others because I had to find the rational approximations of any irrational number in the expression. This was important because in finding the answer because I was working with square roots without a calculator so I had to approximate their values in order to solve the expression. Test Total
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